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Variance

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Anthony Scemama 2021-01-19 10:45:36 +01:00 committed by GitHub
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@ -467,14 +467,24 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
\sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[ \sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[
E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}} E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}
$$ $$
which can be simplified as
$$ \sigma^2(E_L) = \langle E^2 \rangle - \langle E \rangle^2 $$
If the local energy is constant (i.e. $\Psi$ is an eigenfunction of If the local energy is constant (i.e. $\Psi$ is an eigenfunction of
$\hat{H}$) the variance is zero, so the variance of the local $\hat{H}$) the variance is zero, so the variance of the local
energy can be used as a measure of the quality of a wave function. energy can be used as a measure of the quality of a wave function.
*** Exercise (optional)
#+begin_exercise
Prove that :
$$ \sigma^2(E_L) = \langle E^2 \rangle - \langle E \rangle^2 $$
#+end_exercise
*** Exercise *** Exercise
#+begin_exercise #+begin_exercise
Compute a numerical estimate of the variance of the local energy Add the calculation of the variance to the previous code, and
compute a numerical estimate of the variance of the local energy
in a grid of $50\times50\times50$ points in the range in a grid of $50\times50\times50$ points in the range
$(-5,-5,-5) $(-5,-5,-5)
\le \mathbf{r} \le (5,5,5)$ for different values of $a$. \le \mathbf{r} \le (5,5,5)$ for different values of $a$.
@ -492,6 +502,7 @@ r = np.array([0.,0.,0.])
for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]: for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
E = 0. E = 0.
E2 = 0.
norm = 0. norm = 0.
for x in interval: for x in interval:
r[0] = x r[0] = x
@ -503,20 +514,11 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
w = w * w * delta w = w * w * delta
El = e_loc(a, r) El = e_loc(a, r)
E += w * El E += w * El
E2 += w * El*El
norm += w norm += w
E = E / norm E = E / norm
s2 = 0. E2 = E2 / norm
for x in interval: s2 = E2 - E*E
r[0] = x
for y in interval:
r[1] = y
for z in interval:
r[2] = z
w = psi(a, r)
w = w * w * delta
El = e_loc(a, r)
s2 += w * (El - E)**2
s2 = s2 / norm
print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}") print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}")
#+end_src #+end_src